Practicing Success
The absolute maximum value of y = x3 – 3x + 2, 0 ≤ x ≤ 2, is |
4 6 2 0 |
4 |
i) $f(x)=x^3−3x+2;0⩽x⩽2$ $f′(x)=3x^2−3=0⇒x=±1$ critical points are +1 & -1, but we have to find absolute maximum in range [0,2] only. Hence we will reject -1. ii) Evaluating f(x) at critical points & at end points: f(1) = 1 − 3 + 2 = 0 f(0) = 0 − 0 + 2 = 2 f(2) = 8 − 6 + 2 = 4 From above we can see f(x) achieves maximum value of 4 at x = 2 in range x ϵ [0,2] |